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STRUCTURE OF CALCIUM ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy). (August 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism " (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Calcium (Ca) has a total of 24 isotopes, from Ca-34 to Ca-57. There are five stable isotopes (Ca-40, Ca-42, Ca-43, Ca-44 and Ca-46), plus one isotope (Ca-48) with such a long half-life that for all practical purposes it can be considered stable. Calcium also has a cosmogenic isotope, radioactive Ca-41, which has a half-life of 102,000 years. Unlike cosmogenic isotopes that are produced in the atmosphere, Ca-41 is produced by neutron activation of Ca-40. Most of its production is in the upper metre or so of the soil column where the cosmogenic neutron flux is still sufficiently strong. Ca-41 has received much attention in stellar studies because it decays to K-41, a critical indicator of solar-system anomalies. The most stable artificial radioisotope is Ca-45, with a half-life of 163 days. All other isotopes have half-lives of 163 days or less, most under a minute. The least stable is Ca-34 with a half-life shorter than 35 nanoseconds. Ca-40 comprises about 97% of naturally occurring calcium. Ca-40 is also one of the daughter products of K-40 decay, along with Ar-40. WHY Ca-40, Ca-42 Ca-43 Ca-44 , Ca-46 AND Ca-48 ARE STABLE NUCLIDES ''' After a careful analysis of the structure of atomic nuclei I discovered that the beta decay is due to the fact that in unstable nuclei there exist single horizontal pn bonds of weak binding energy leading to the beta decay. For example in my paper STRUCTURE AND BINDING OF H3 AND He3 using the diagram of the structure of the H3 one sees that it is unstable because the two neutrons make single np bonds, while the He3 is stable because the one neutron between the two protons makes two np bonds per neutron. On the other hand the pp repulsions of long range lead to the instability when we have a small number of pn bonds per nucleon. For understanding better why the calcium with S = 0 having an even number of neutrons (from Ca-40 to Ca-48) contains stable structures you can see the Fig. 7a of my published paper “Nuclear structure is governed by the fundamental laws of electromagnetism”. However for describing the stable structure of Ca-43 with S = -7/2 (odd number of neutrons) you can see the following diagram of Ca-43 . In this stable structure the core is not the parallelepiped of Mag-24 with S=0. Here the core consists of a parallelepiped with the five horizontal planes which exist from the -HP1 to the -HP5 giving S = -3 . Here you also see that the deuterons like the p16n16, n17p17, and p18n18 give S = -1 . Whereas the two vertical pn systems with strong vertical bonds like the n19p19 and p20n20 give S=0. Note that the 19p19 is not shown, because it exists in front of the p3 and n5. Also the p20n20 is not shown because it exists behind the n4 and p6. It is of interest to notice that the p19 with the p7 and also the p20 with the p2 form blank positions for receiving the extra n21(+1/2) at +HP4 and the extra n22(-1/2) at -HP1. They also are not shown because they are in front of p7 and behind the p2 respectively. You can see only the n23(+1/2) which fills the blank position formed by the p18 and p7. Thus the total spin is given by S = -3 -1 +1/2 -1/2 +1/2 = -7/2 In other words the stability of the above nuclides is due to the fact that the extra neutrons make two bonds per neutron able to overcome the pp and nn repulsions. '''DIAGRAM OF STABLE Ca-43 WITH S =-7/2 The structure has five horizontal planes of opposite spins. For example the first horizontal plane (-HP1) has nucleons of negative spins. Here the n19, p19 and n21 are not shown because they are in front of p3, n5 and p7 respectively. Also the n22, p20 and n20 are not shown because they are behind the p2, n4 and p6 respectively ' ' ' p15.....n10......p10' ' -HP5 n15.......p9.......n9 ' ' n14.......p8........n8 ' ' +HP4 p14......n7........ p7......n23 ' ' p13.......n6........p6........n18' ' -HP3 n13.......p5........n5.......p18 ' ' n12.......p4........n4........p17' ' +HP2 p12......n3........p3.......n17 ' ' p11......n2........p2.......n16' ' -HP1 n11.....p1.......n1........p16 ' ' NUCLEAR STRUCTURE OF Ca-50, Ca-52, Ca-54 AND Ca-56 WITH S =0 ' Since the protons of the alpha particles existing outside the parallelepiped of Mg-24 cannot form more blank positions than the number of 8 we conclude that the structure of the above nuclides is based on the structure of Ca-48 in which the more extra neutrons than those of Ca-48 make single bonds leading to the decay. For example the Ca-56 with S=0 has 8 more extra neutrons of opposite spins than those of Ca-48 with S=0. ' ' NUCLEAR STRUCTURE OF Ca-38, Ca-36, AND Ca-34 WITH S = 0 In the absence of neutrons with opposite spins we get the structure of the above nuclides based on the structure of Ca-40 with S=0. For example in the structure of Ca-34 we have 6 absent neutrons of opposite spins ' ' NUCLEAR STRUCTURE OF Ca-45, Ca-47, Ca-49 Ca-51, Ca-53, Ca-55 AND Ca-57 ' After a careful analysis I found that the structure of the above nuclides is based on the structure of Ca-43. For example the Ca-47 with S = -7/2 has 4 extra neutrons of opposite spins more than those of Ca-43 with S=-7/2 . In the case of Ca-49 with S=-3/2 we conclude that it has 4 more extra neutrons of positive spins and 2 more extra neutrons of opposite spins giving S=0. That is S = -7/2 + 4(+1/2) + 0 = -3/2. Also in the case of Ca-57 with S=-5/2 we conclude that it has 2 more extra neutrons of positive spins and 12 more extra neutrons of opposite spins giving S=0. ' ''' '''NUCLEAR STRUCTURE OF Ca-41, Ca-39, Ca-37 AND Ca-35 Here the structure of Ca-41 with S =-7/2 is based on the structure of Ca-43, because in the absence of two neutrons of opposite spins we get the structure of Ca-41 with the same S=-7/2. However for describing the structure of Ca-39 with S = +3/2 we see that it is based on the similar structure of Ca -43. In this new case the p16(-1/2) goes from the -HP1 to +HP4 for making a deuteron with the n23(+1/2). But his change of spin gives S =+1. So the new total spin of the Ca-43 is S = -5/2. Under this new structure when all nucleons change the spins we will get a total spin S = +5/2 . For example now we have +HP1, -HP2, +HP3, -HP4 and +HP5 giving S = +5/2. Thus, in the absence of two neutrons of positive spins like the n15 and n16 we get the structure of Ca-39 with S=+3/2. That is S = +5/2 -2(+1/2) = +3/2 Moreover in the structure of Ca-37 with S =+3/2 we have two more absent neutrons of opposite spins than those of Ca-39. Also the structure of Ca-35 with S =+1/2 is based on the structure of Ca-37 with S=+3/2. In this case we have two absent neutrons of positive spins more than those of Ca-37 with S=+3/2. That is S = +3/2 -2(+1/2) = +1/2. Category:Fundamental physics concepts